Chi-Square With Base Hypothesis That A-Level Coursework

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Total Length: 731 words ( 2 double-spaced pages)

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23343849

73

0.35009171

35-54

88.40378549

82

0.46387684

55+

81.36277603

93

1.66445872

2 = 11.39

This value does exceed the critical ?2 value for df = 2 at ? = 0.05. Therefore, we can assume that one of the observed values is significantly different from the expected value for that group. Without post-hoc pairwise tests it is impossible to say exactly which group is different. We can make an educated guess, however, that the proportion of 55+ shoppers in store a is statistically different from what would be expected by chance.

3. Collapse the response categories in the following table so that it meets the assumption of the Chi-square test, then perform the test.

Ownership (Collapsed)

Education

Owners

Non-owners

Some High School or Below

5

17

High School graduate

30

25

22

26

Post-Baccalaureate

5

7

Total

62

75

2 = 6.49. This does not exceed the critical ?2 value for df = 3, so we cannot assume that there is any significant difference between the observed counts of home ownership by educational level and those expected by chance.

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4. A ?2 test to determine whether the sample is significantly different from the expected distribution would be most appropriate. The data in this case yield ?2 = 2.51, below the critical value cutoff for ? = 0.05. We can assume that the sample is NOT significantly different from the general population.

5. In this data, there appears to be no gender-based difference between the way in which people commute to work. Since ?2 = 7.715, df = 3 and p > 0.1, this data fails to exceed the critical ?2 value for the analysis, 7.81.

6. To test this hypothesis I will use a t-test of means. Critical t-value for ? = 0.05 is (conservatively!) 1.98.

T = (M1 - M2) / SE (M1 - M2)

SE = ( (var1/n1) + (var2/n2) = ( (0.52/100) + (0.62/64) = 0.09

T = 1.....

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